Question

Tính nồng độ mol các ion Al³⁺,SO₄^2-trong dung dịch Al₂(SO₄)₃ 25% (D = 1,386g/ml)?

Answer 1

Giả sử có 1 lít dung dịch Al₂(SO₄)₃ 25%

\(m_{dd}=1000.1,386=1386\left(g\right)\)

\(m_{Al_2\left(SO_4\right)_3}=\dfrac{1386.25}{100}=346,5\left(g\right)\Rightarrow n_{Al_2\left(SO_4\right)_3}=\dfrac{346,5}{342}=\dfrac{77}{76}\left(mol\right)\)

   Al₂(SO₄)₃ --> 2Al³⁺ + 3SO₄^2-

       \(\dfrac{77}{76}\)------->\(\dfrac{77}{38}\)------>\(\dfrac{231}{76}\)

\(\left\{{}\begin{matrix}C_{Al^{3+}}=\dfrac{\dfrac{77}{38}}{1}=\dfrac{77}{38}M\\C_{SO_4^{2-}}=\dfrac{\dfrac{231}{76}}{1}=\dfrac{231}{76}M\end{matrix}\right.\)